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For a harmonic function $u(x)$, on domain $\Omega$ where $x \in \Omega \subset \Bbb R^n $, how to show that $$ u(x) = \frac{1}{\omega_n R^{n-1}}\int_{\partial B_R(x)} u(\sigma) d\sigma$$ where $\omega_n$ is the area of the unit sphere $\partial B_1(x)$.
I am looking for simple case $n=2$ centered at $0$. An argument in my note uses the definition $\displaystyle g(r) \overset{(def)}{=} \frac{1}{2\pi r }\int_{\partial B_r(0)} u(\sigma) d\sigma$ and shows that $g'(r) = 0$. I don't see how does it prove above? I think it proves $g(r) = \text{constant}$.

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  • $\begingroup$ You are right, but what is that constant? $\endgroup$
    – user99914
    Commented Oct 30, 2013 at 5:08
  • $\begingroup$ @John is it supposed to be $u(0)$ ?? $\endgroup$ Commented Oct 30, 2013 at 5:09
  • $\begingroup$ Note that $g(r)$ is the average of $g$ over the circle of radius $r$. If $g$ is continuous, $g(r)$ should goes to $g(0)$. Of course one needs to prove that, though the proof is not difficult. $\endgroup$
    – user99914
    Commented Oct 30, 2013 at 5:53

1 Answer 1

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Hint: consider $\displaystyle\lim_{r\to 0}g(r)$.

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  • $\begingroup$ all right, is $g(r) = g(0)$ ?? $\endgroup$ Commented Oct 30, 2013 at 5:11
  • $\begingroup$ $g$ is not defined at $0$, but its limit at $0$ exists. $\endgroup$
    – detnvvp
    Commented Oct 30, 2013 at 5:12
  • $\begingroup$ could you be a bit explicit in you answer?? I am not getting it :(( $\endgroup$ Commented Oct 30, 2013 at 5:14
  • $\begingroup$ Ok, since $g'=0$, you have that $g$ is a constant. Therefore, $\displaystyle g(r)=\lim_{s\to 0}g(s)$. All you have to do is prove that $\displaystyle\lim_{s\to 0}g(s)=u(0)$. $\endgroup$
    – detnvvp
    Commented Oct 30, 2013 at 5:16
  • $\begingroup$ i see ...${{{}}}$ $\endgroup$ Commented Oct 30, 2013 at 5:18

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